Let be given the functions: $\ \forall z \in \mathbb{C}\displaystyle{f_{1}(z) = \frac{1}{(z-1)^{3}},(z\neq 1); f_{2}= \cos\left(\frac{1}{z}\right)}; (z\neq 0)$ and let $U$ be an open disc with radius $R$ around $z_{0}=1$ for $f_{1}$ and $z_{0}=0$ for $f_{2}$.

Onto which set $V\subset \mathbb{C}$ is the unit disc mapped? How many times do $f_{1}, f_{2}$ take each value $w\in V$ on $U\backslash \{z_{0}\}$?

$f_1$ has a pole of order $3$ at $z_{0}$ , and $\cos(1/z)$ has an essential singularity at $z_{0}$. $f_{1}$ takes each value $w$ $3$ times and $f_{2}$ once. Since it is an essential singularity in $f_2$, that means the map of the disc will be dense and thus onto whole $\mathbb{C}$. For $f_{2}$ since it has a pole of order $3$, the map will not be dense in $\mathbb{C}$.

Can one say onto which set exactly the disc with radius $R$ is mapped onto? How?

1 Answer
1

This looks like a homework problem, if so could you use the "homework" tag? I shall give a few hints.

For $f_1$, consider the image of the disk of radius $R$ around $0$ under the map $z\mapsto z^3$. Then apply the transformation $z\mapsto 1/z$ to this image.

For $f_2$, consider the image of a neighbourhood of the essential singularity at $\infty$ for the map $\cos$. It is true that this is the entire complex plane, but your reasoning is not correct, as a holomorphic function with an essential singularity can omit a value of the complex plane.

Because $\cos$ is $2\pi$-periodic, it takes every value infinitely often (you stated it takes every value only once, but that is false). But we will also see this directly.

Recall the definition of $\cos$ as $\cos(z) = \frac{e^{iz}+e^{-iz}}{2}$. So $\cos(z) = g(e^{iz})$, where $g(z) = \frac{1}{2}\left(z+\frac{1}{z}\right)$. Use the fact that $g$ takes every complex value, that the poles of $g$ are at $0$ and $\infty$, and that $e^z$ takes every nonzero complex value infinitely often, to conclude that $\cos$ takes every complex value infinitely often.